Question

A standard flashlight battery can deliver about 2.0 W-h of energy before it runs down. (a) If a battery costs US \(0.80, what is the cost of operating a 100 W lamp for 8.0h using batteries? (b) What is the cost of energy is provided at the rate of US \)0.6 per kilowatt-hour?

Short answer

1. The cost of operating a 100W lamp for 8.0h using batteries is$\$3.2\times {10}^2$.
2. The cost of energy provided at the rate of$US\$0.6kilowatt-hour$is 4.8cents.

Step-by-step solution

 Step 1: The given data:

• The energy delivered by the flashlight battery, $E=2.0W.h$
• Input power of the lamp,$P_{i}n=100W$
• Time of operation of the battery,$t=8.0h$
• Cost of one battery, $Cos{t}_{onebattery}=US\$0.8$
• Cost of energy, $Cos{t}_{energyunit}=US\$0.06kilowatthour$

Understanding the concept of power:

A battery operating in a given circuit consumes some amount of energy within a given time and that is given as the power of the battery. Thus, in the given problem, a lamp is operating with the use of some batteries and consumes a desired amount of energy to stay lit up. As you know, one unit of energy consumption is considered one kilowatt per hour in a household, so using the total energy, you can get the cost rate of energy consumption by the lamp.

Formulae:

The power of a due to energy consumption,

$P=\frac{E}{t}$ ….. (i)

Where, E is the amount of energy consumed, t is the time of consumption or the battery operation of the battery.

The number of batteries in a lamp,

$N=\frac{P_{in}}{P_{out}}$ ..........(ii)

Where, P_in is the input power of the lamp and P_out is the output power delivered by the lamp.

(a) Calculation of the operating the lamp:

Here, the cost of running the lap is given by the cost it takes in running the number of batteries it uses.

Now, the output power or the power delivered by one battery can be given using the given data in equation (i) as follows:

$$\begin{gathered} P_{out}=\frac{2.0W.h}{8.0h} \\ =0.25W \end{gathered}$$

Thus, the number of batteries operating for the lap is given using equation (ii) as follows:

$$\begin{gathered} N=\frac{100W}{0.25W} \\ =400batteries \end{gathered}$$

Now, the cost of operating a 100W lamp for 8.0husing batteries is given as follows:$$\begin{gathered} total\cos t=N\times Cos{t}_{onebattery} \\ =400US\$0.8 \\ =\$3200 \\ =\$3.2\times {10}^2 \end{gathered}$$

Hence, the required cost of the operating lamp is $3.2*10²

(b) Calculation of the cost of total energy

The total energy consumption by the lamp is given using the data in equation (i) as follows:

$\begin{array}{l}E=P.t\\ =\left(100w\left)\right(8.0h\right)\\ =800W.h\end{array}$

If one kilowatt hour energy costs US$0.06, then the cost of total energy is given as follows:

$$\begin{gathered} Total\cos t=\frac{800kW.h}{{10}^3}\times US\$0.6 \\ =\$0.048 \\ =4.8cents \end{gathered}$$

Hence, the value of the required cost is 4.8cents.